DEM 7223 - Event History Analysis - Comparing Survival Times Between Groups
Corey S. Sparks, PhD
The University of Texas at San Antonio18 August, 2020
Product Limit Estimation
Kaplan and Meier (1958) derived an estimator of the survivorship function for a sample of censored and uncensored cases.
The method figured survival to a time, t, is the product of the survival from all previous time points.
i.e. you can only get to time 3, if you survive time 1 and time 2, etc
We can write this as
\[\hat{S(t)} = (1-\hat{p(t_1)})(1-\hat{p(t_2)}) \dots (1-\hat{p(t_j)})\]
Unlike the life-table and discrete time methods of estimating survival, which lumped time into discrete periods, K-M uses the information contained in the actual duration.
Each K-M interval begins with a single event time, and ends just prior to the next event time.
Kaplan-Meier Estimation
The K-M estimator is written: \[\hat{S(t)} = \prod_{t_i \leqslant t}\frac{n_i - d_i}{n_i} = \prod_{t_i \leqslant t} \left[1- \frac{d_i}{Y_i} \right]\]
Where the \(t_i\) are the ranked survival times, \(n_i\) is the number of individuals at risk at each time, \(t_i\) is time, \(d_i\) is the number of events at each time.
When censoring is present, you define \(n_i\) by subtracting out the number of censored cases at that particular time.
t1<-data.frame(Time_Interval=c("[0,1)", "[1,2)", "[2,4)", "[4, 5)", "[5, 6)", "[6+)"),
time=c(0, 1, 2, 4, 5, 6),
n=c(100,NA, NA, NA, NA, NA),
d=c(15, 5, 1, 2, 5, 2),
c=c(0,2,5,2,0,2),
prob=c(1,NA, NA, NA, NA, NA),
St=c(1,NA, NA, NA, NA, NA))
for (i in 2:6){
t1$n[i]<-t1$n[i-1]- t1$d[i-1] -t1$c[i-1]
t1$prob[i]<-1-(t1$d[i]/t1$n[i])
}
t1$St<-cumprod(t1$prob)
t1%>%
kable()%>%
column_spec(1:4, border_left = T, border_right = T)%>%
kable_styling()| Time_Interval | time | n | d | c | prob | St |
|---|---|---|---|---|---|---|
| [0,1) | 0 | 100 | 15 | 0 | 1.0000000 | 1.0000000 |
| [1,2) | 1 | 85 | 5 | 2 | 0.9411765 | 0.9411765 |
| [2,4) | 2 | 78 | 1 | 5 | 0.9871795 | 0.9291101 |
| [4, 5) | 4 | 72 | 2 | 2 | 0.9722222 | 0.9033015 |
| [5, 6) | 5 | 68 | 5 | 0 | 0.9264706 | 0.8368823 |
| [6+) | 6 | 63 | 2 | 2 | 0.9682540 | 0.8103146 |
t1%>%
ggplot()+
geom_step(aes(x=time, y=St,color="a"))+
geom_step(aes(x=time, y=prob, color="b"))+
scale_colour_identity(guide="legend")+
scale_colour_manual(name = 'Function Type',
values =c('a'='red','b'='green'),
labels = c('S(t)','p(t)'))+
xlab("Time")+ylab("Probability")+
ggtitle("Kaplan - Meier Survival and Probability Functions")## Scale for 'colour' is already present. Adding another scale for 'colour',
## which will replace the existing scale.
K-M and the hazard function
The exact estimate of the K-M hazard function actually depends on the width of the time interval at each observed time point.
\[\hat{h(t_j)} = \frac{\hat{p_{KM}(t_j)}}{\text{width}_j}\] and you can get this estimate from the muhaz library in R.
Variance in the K-M Estimates
Since the K-M survival function is a statistical estimate, it also has uncertainty to it. To measure the uncertainty, or variance in the estimate, the traditional method is to use the Greenwood formula.
\[\text{Var}(S(t)) = \hat{S(t)}^2 \sum_{t_i \leqslant t} \frac{d_i}{n_i(n_i-d_i)}\] and standard error equal to \(s.e.(S(t))= \sqrt{\text{Var}(S(t))}\)
If we have a standard error of the survival function, the assuming the sampling distribution of the survival function is normal, we can calculate a normal confidence interval for the survival function at each time:
\[c.i.(S(t)) = \hat{S(t)} \pm z_{1-\alpha/2} * s.e.(S(t))\]
where \(z\) is the standard normal variate corresponding to the the \(1-\alpha/2\) level of confidence.
Estimating the cumulative hazard function.
If we have estimates of \(\hat{h(t_j)}\), we can either calculate the value of the cumulative hazard function using the relationship among survival function:
\[H(t) = -\text{log } S(t)\] or use the Nelson-Aalen estimator:
\[\hat{H(t)} = \sum_{t_i \leqslant t} \frac{d_i}{Y_i}\]
Comparing Survival curves between groups
Often in our data there are distinct groups for which we want to compare survival patterns * e.g. treatment vs. control group in clinical trial * e.g. proportion of women without a first birth by education category
To do this we typically construct a variable that represents an identifier for members of reach group. This can be referred to as an indicator, or class, variable.
This is the same process as doing a two sample test for a regular outcome, such as a t-test or chi square test.
One limitation of survival data is that they are typically skewed, meaning their distribution is not symmetrical
This means that many traditional hypothesis test (t-test, z-test) for comparison of central tendency are no appropriate
Instead we use a variety of non-parametric methods
Simply meaning that these tests are not dependent on the shape of the distribution or on the parameters of the distribution
Also, due to censoring, traditional distributional parameters like the mean are less meaningful, so tests on said parameters would be incorrect
Graphical methods of comparison
The first stop on our examination of between group comparison is the inter-ocular traumatic test
You may not be familiar with this test, but in general, if you look at a plot, and if you think there’s a difference, say between two lines, there usually is, and many times the human eye is a more discerning test than anything.
## Warning: `data_frame()` is deprecated as of tibble 1.1.0.
## Please use `tibble()` instead.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.
So, plot the survival curves, with confidence intervals, your eye can usually detect if there is a difference
Under traditional statistics thinking, if the confidence intervals of the two curves overlap for their entire lengths, then the two groups are equivalent, if the confidence interval for the curves do not overlap at ANY point along the curve, they are different, simple, no?
The statistical way of doing this, beyond looking at things, is the realm of Mantel-Haenszel test. R implements this test for 2 or k groups using the survdiff() function. It uses the method of Harrington and Fleming (1982) which can weight the difference in survival curves flexibly, giving more or less weight to earlier or later survival times.
The classic Mantel-Haneszel test is just a \(\chi^2\) test for independence.
At each time point, \(t_i\), consider the following table for 2 groups:
M-H Test
If we sum the differences between the observed and expected failures across all time points,we arrive with:
Using 1 group as the basis: \[e_{i1} = \frac{n_{i1}d_{i1}}{n_i}\] is the expected number of failures. The general form of the test is then:
\[Q = \frac{\sum_{i} w_i (d_{i1} - e_{i1})^2}{\sum_i w_i v_{i1}}\] Where \(v_{i1}\) is the variance in the number of events in group 1. This test follows a \(\chi^2\) distribution with 1 degree of freedom.
The value of \(w_i\) allows great flexibility for these tests, and is called the weight function at time \(i\) This allows the analyst to specify how much you want to weight the difference in survival at a particular time point.
This testing logic also extends to k-groups, so instead of doing an ANOVA test, you would do the k-group test following this method.
Example
This example will illustrate how to test for differences between survival functions estimated by the Kaplan-Meier product limit estimator. The tests all follow the methods described by Harrington and Fleming (1982) Link.
The first example will use as its outcome variable, the event of a child dying before age 1. The data for this example come from the model.data Demographic and Health Survey for 2012 children’s recode file. This file contains information for all births in the last 5 years prior to the survey.
The second example, we will examine how to calculate the survival function for a longitudinally collected data set. Here I use data from the ECLS-K. Specifically, we will examine the transition into poverty between kindergarten and fifth grade.
#load libraries
library(haven)
library(survival)
library(car)## Loading required package: carData##
## Attaching package: 'car'## The following object is masked from 'package:dplyr':
##
## recodelibrary(muhaz)
model.dat<-read_dta("https://github.com/coreysparks/data/blob/master/ZZKR62FL.DTA?raw=true")
model.dat<-zap_labels(model.dat)Event - Infant Mortality
In the DHS, they record if a child is dead or alive and the age at death if the child is dead. This can be understood using a series of variables about each child.
If the child is alive at the time of interview, then the variable B5==1, and the age at death is censored.
If the age at death is censored, then the age at the date of interview (censored age at death) is the date of the interview - date of birth (in months).
If the child is dead at the time of interview,then the variable B5!=1, then the age at death in months is the variable B7. Here we code this:
model.dat$death.age<-ifelse(model.dat$b5==1,
((((model.dat$v008))+1900)-(((model.dat$b3))+1900))
,model.dat$b7)
#censoring indicator for death by age 1, in months (12 months)
model.dat$d.event<-ifelse(is.na(model.dat$b7)==T|model.dat$b7>12,0,1)
model.dat$d.eventfac<-factor(model.dat$d.event); levels(model.dat$d.eventfac)<-c("Alive at 1", "Dead by 1")
table(model.dat$d.eventfac)##
## Alive at 1 Dead by 1
## 5434 534Comparing Two Groups
We will now test for differences in survival by characteristics of the household. First we will examine whether the survival chances are the same for children in relatively high ses (in material terms) households, compared to those in relatively low-ses households.
This is the equivalent of doing a t-test, or Mann-Whitney U test for differences between two groups.
library(survminer)## Loading required package: ggpubrmodel.dat$highses<-Recode(model.dat$v190, recodes ="1:3 = 0; 4:5=1; else=NA")
fit1<-survfit(Surv(death.age, d.event)~highses, data=model.dat)
fit1## Call: survfit(formula = Surv(death.age, d.event) ~ highses, data = model.dat)
##
## n events median 0.95LCL 0.95UCL
## highses=0 4179 362 NA NA NA
## highses=1 1789 172 NA NA NAggsurvplot(fit1, xlim=c(0,12), conf.int=T, title="Survival Function for Infant Mortality - Low vs. High SES Households", ylim=c(.8, 1))
summary(fit1)## Call: survfit(formula = Surv(death.age, d.event) ~ highses, data = model.dat)
##
## highses=0
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 0 4179 134 0.968 0.00273 0.963 0.973
## 1 3992 18 0.964 0.00290 0.958 0.969
## 2 3914 28 0.957 0.00316 0.951 0.963
## 3 3808 24 0.951 0.00337 0.944 0.957
## 4 3709 10 0.948 0.00346 0.941 0.955
## 5 3625 15 0.944 0.00359 0.937 0.951
## 6 3520 20 0.939 0.00376 0.931 0.946
## 7 3414 14 0.935 0.00389 0.927 0.943
## 8 3325 21 0.929 0.00407 0.921 0.937
## 9 3238 17 0.924 0.00422 0.916 0.932
## 10 3159 3 0.923 0.00424 0.915 0.932
## 11 3090 10 0.920 0.00433 0.912 0.929
## 12 3015 48 0.906 0.00475 0.896 0.915
##
## highses=1
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 0 1789 75 0.958 0.00474 0.949 0.967
## 1 1698 8 0.954 0.00498 0.944 0.963
## 2 1659 9 0.948 0.00524 0.938 0.959
## 3 1615 14 0.940 0.00564 0.929 0.951
## 4 1573 15 0.931 0.00604 0.919 0.943
## 5 1536 7 0.927 0.00622 0.915 0.939
## 6 1501 6 0.923 0.00638 0.911 0.936
## 7 1466 4 0.921 0.00648 0.908 0.934
## 8 1430 5 0.918 0.00662 0.905 0.931
## 9 1383 6 0.914 0.00679 0.900 0.927
## 10 1348 2 0.912 0.00684 0.899 0.926
## 11 1315 3 0.910 0.00693 0.897 0.924
## 12 1288 18 0.897 0.00746 0.883 0.912Gives us the basic survival plot.
Next we will use survtest() to test for differences between the two or more groups. The survdiff() function performs the log-rank test to compare the survival patterns of two or more groups.
#two group compairison
survdiff(Surv(death.age, d.event)~highses, data=model.dat)## Call:
## survdiff(formula = Surv(death.age, d.event) ~ highses, data = model.dat)
##
## N Observed Expected (O-E)^2/E (O-E)^2/V
## highses=0 4179 362 374 0.401 1.37
## highses=1 1789 172 160 0.940 1.37
##
## Chisq= 1.4 on 1 degrees of freedom, p= 0.2In this case, we see no difference in survival status based on household SES.
How about rural vs urban residence?
library(dplyr)
library(car)
model.dat<-model.dat%>%
mutate(rural = car::Recode(v025, recodes ="2 = '0rural'; 1='1urban'; else=NA", as.factor = T))
fit2<-survfit(Surv(death.age, d.event)~rural, data=model.dat, conf.type = "log")
fit2## Call: survfit(formula = Surv(death.age, d.event) ~ rural, data = model.dat,
## conf.type = "log")
##
## n events median 0.95LCL 0.95UCL
## rural=0rural 4138 346 NA NA NA
## rural=1urban 1830 188 NA NA NAsummary(fit2)## Call: survfit(formula = Surv(death.age, d.event) ~ rural, data = model.dat,
## conf.type = "log")
##
## rural=0rural
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 0 4138 130 0.969 0.00271 0.963 0.974
## 1 3959 15 0.965 0.00286 0.959 0.971
## 2 3879 29 0.958 0.00314 0.952 0.964
## 3 3775 24 0.952 0.00336 0.945 0.958
## 4 3682 15 0.948 0.00349 0.941 0.955
## 5 3590 13 0.944 0.00360 0.937 0.951
## 6 3491 17 0.940 0.00375 0.932 0.947
## 7 3382 13 0.936 0.00387 0.929 0.944
## 8 3299 19 0.931 0.00404 0.923 0.939
## 9 3208 17 0.926 0.00419 0.918 0.934
## 10 3126 3 0.925 0.00422 0.917 0.933
## 11 3060 9 0.922 0.00430 0.914 0.931
## 12 2990 42 0.909 0.00469 0.900 0.918
##
## rural=1urban
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 0 1830 79 0.957 0.00475 0.948 0.966
## 1 1731 11 0.951 0.00506 0.941 0.961
## 2 1694 8 0.946 0.00528 0.936 0.957
## 3 1648 14 0.938 0.00566 0.927 0.949
## 4 1600 10 0.932 0.00592 0.921 0.944
## 5 1571 9 0.927 0.00615 0.915 0.939
## 6 1530 9 0.922 0.00637 0.909 0.934
## 7 1498 5 0.918 0.00650 0.906 0.931
## 8 1456 7 0.914 0.00668 0.901 0.927
## 9 1413 6 0.910 0.00683 0.897 0.924
## 10 1381 2 0.909 0.00689 0.895 0.922
## 11 1345 4 0.906 0.00700 0.893 0.920
## 12 1313 24 0.890 0.00764 0.875 0.905ggsurvplot(fit2, xlim=c(0,12), ylim=c(.8, 1), conf.int=T, title="Survival Function for Infant mortality - Rural vs Urban Residence")
Two- sample test
survdiff(Surv(death.age, d.event)~rural, data=model.dat)## Call:
## survdiff(formula = Surv(death.age, d.event) ~ rural, data = model.dat)
##
## N Observed Expected (O-E)^2/E (O-E)^2/V
## rural=0rural 4138 346 371 1.67 5.55
## rural=1urban 1830 188 163 3.79 5.55
##
## Chisq= 5.6 on 1 degrees of freedom, p= 0.02prop.table(table(model.dat$d.event, model.dat$rural), margin = 2)##
## 0rural 1urban
## 0 0.91638473 0.89726776
## 1 0.08361527 0.10273224chisq.test(table(model.dat$d.event, model.dat$rural))##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: table(model.dat$d.event, model.dat$rural)
## X-squared = 5.4595, df = 1, p-value = 0.01946Which shows a statistically significant difference in survival between rural and urban children, with rural children showing lower survivorship at all ages.
We can also compare the 95% survival point for rural and urban residents
quantile(fit2, probs=.05)## $quantile
## 5
## rural=0rural 4
## rural=1urban 2
##
## $lower
## 5
## rural=0rural 3
## rural=1urban 0
##
## $upper
## 5
## rural=0rural 6
## rural=1urban 3We can also calculate the hazard function for each group using the kphaz.fit function in the muhaz library.
haz2<-kphaz.fit(model.dat$death.age, model.dat$d.event, model.dat$rural)
haz2## $time
## [1] 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5 0.5 1.5 2.5
## [16] 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5
##
## $haz
## [1] 0.0038314014 0.0075716497 0.0064378810 0.0041197145 0.0036742403
## [6] 0.0049398949 0.0038930677 0.0058337248 0.0053636408 0.0009670736
## [11] 0.0029830591 0.0143373352 0.0064084477 0.0047764602 0.0086460188
## [16] 0.0063018400 0.0058154299 0.0059338979 0.0033648177 0.0048901211
## [21] 0.0042929999 0.0014588521 0.0030184080 0.0185979202
##
## $var
## [1] 9.786715e-07 1.976995e-06 1.727037e-06 1.131544e-06 1.038557e-06
## [6] 1.435586e-06 1.165912e-06 1.791334e-06 1.692389e-06 3.117631e-07
## [11] 9.887982e-07 4.894885e-06 3.733606e-06 2.852025e-06 5.339922e-06
## [16] 3.971474e-06 3.757836e-06 3.912526e-06 2.264454e-06 3.416455e-06
## [21] 3.071735e-06 1.064171e-06 2.277779e-06 1.441308e-05
##
## $strata
## [1] 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2plot(y=haz2$haz[1:12], x=haz2$time[1:12], col=1, lty=1, type="s")
lines(y=haz2$haz[13:24], x=haz2$time[13:24], col=2, lty=1, type="s")
This may be suggestive that children in urban areas may live in poorer environmental conditions.
k- sample test
Next we illustrate a k-sample test. This would be the equivalent of the ANOVA if we were doing ordinary linear models.
In this example, I use the v024 variable, which corresponds to the region of residence in this data. Effectively we are testing for differences in risk of infant mortality by region.
table(model.dat$v024, model.dat$d.eventfac)##
## Alive at 1 Dead by 1
## 1 2229 181
## 2 1435 141
## 3 631 94
## 4 1139 118fit3<-survfit(Surv(death.age, d.event)~v024, data=model.dat)
fit3## Call: survfit(formula = Surv(death.age, d.event) ~ v024, data = model.dat)
##
## n events median 0.95LCL 0.95UCL
## v024=1 2410 181 NA NA NA
## v024=2 1576 141 NA NA NA
## v024=3 725 94 NA NA NA
## v024=4 1257 118 NA NA NA#summary(fit3)
#quantile(fit3, probs=.05)
ggsurvplot(fit3,conf.int = T, risk.table = F, title = "Survivorship Function for Infant Mortality", xlab = "Time in Months", xlim = c(0,12), ylim=c(.8, 1))
survdiff(Surv(death.age, d.event)~v024, data=model.dat)## Call:
## survdiff(formula = Surv(death.age, d.event) ~ v024, data = model.dat)
##
## N Observed Expected (O-E)^2/E (O-E)^2/V
## v024=1 2410 181 215.5 5.52534 9.43344
## v024=2 1576 141 141.9 0.00537 0.00745
## v024=3 725 94 62.9 15.33021 17.70233
## v024=4 1257 118 113.7 0.16401 0.21218
##
## Chisq= 21.4 on 3 degrees of freedom, p= 9e-05Which shows significant variation in survival between regions. The biggest difference we see is between region 3 green) and region 1 (black line) groups.
Lastly, we examine comparing survival across multiple variables, in this case the education of the mother (secedu) and the rural/urban residence rural:
model.dat<-model.dat%>%
mutate(secedu=Recode(v106, recodes ="2:3 = 1; 0:1=0; else=NA"))
table(model.dat$secedu, model.dat$d.eventfac)##
## Alive at 1 Dead by 1
## 0 4470 430
## 1 964 104fit4<-survfit(Surv(death.age, d.event)~rural+secedu, data=model.dat)
#summary(fit4)
ggsurvplot(fit4,conf.int = T, risk.table = F, title = "Survivorship Function for Infant Mortality", xlab = "Time in Months", xlim = c(0,12), ylim=c(.8, 1))
#plot(fit4, ylim=c(.85,1), xlim=c(0,12), col=c(1,1,2,2),lty=c(1,2,1,2), conf.int=F)
#title(main="Survival Function for Infant Mortality", sub="Rural/Urban * Mother's Education")
#legend("topright", legend = c("Urban, Low Edu","Urban High Edu ", "Rural, Low Edu","Rural High Edu " ), col=c(1,1,2,2),lty=c(1,2,1,2))
# test
survdiff(Surv(death.age, d.event)~rural+secedu, data=model.dat)## Call:
## survdiff(formula = Surv(death.age, d.event) ~ rural + secedu,
## data = model.dat)
##
## N Observed Expected (O-E)^2/E (O-E)^2/V
## rural=0rural, secedu=0 3707 308 333.3 1.92685 5.22186
## rural=0rural, secedu=1 431 38 37.5 0.00632 0.00693
## rural=1urban, secedu=0 1193 122 107.1 2.07801 2.64671
## rural=1urban, secedu=1 637 66 56.1 1.76227 2.00529
##
## Chisq= 5.9 on 3 degrees of freedom, p= 0.1Which shows a marginally significant difference between at least two of the groups, in this case, I would say that it’s most likely finding differences between the Urban, low Education and the Rural low education, because there have the higher ratio of observed vs expected.
Survival analysis using survey design
This example will cover the use of R functions for analyzing complex survey data. Most social and health surveys are not simple random samples of the population, but instead consist of respondents from a complex survey design. These designs often stratify the population based on one or more characteristics, including geography, race, age, etc. In addition the designs can be multi-stage, meaning that initial strata are created, then respondents are sampled from smaller units within those strata. An example would be if a school district was chosen as a sample strata, and then schools were then chosen as the primary sampling units (PSUs) within the district. From this 2 stage design, we could further sample classrooms within the school (3 stage design) or simply sample students (or whatever our unit of interest is).
A second feature of survey data we often want to account for is differential respondent weighting. This means that each respondent is given a weight to represent how common that particular respondent is within the population. This reflects the differenital probability of sampling based on respondent characteristics. As demographers, we are also often interested in making inference for the population, not just the sample, so our results must be generalizable to the population at large. Sample weights are used in the process as well.
When such data are analyzed, we must take into account this nesting structure (sample design) as well as the respondent sample weight in order to make valid estimates of ANY statistical parameter. If we do not account for design, the parameter standard errors will be incorrect, and if we do not account for weighting, the parameters themselves will be incorrect and biased.
In general there are typically three things we need to find in our survey data code books: The sample strata identifier, the sample primary sampling unit identifier (often called a cluster identifier) and the respondent survey weight. These will typically have one of these names and should be easily identifiable in the code book.
Statistical software will have special routines for analyzing these types of data and you must be aware that the diversity of statistical routines that generally exists will be lower for analyzing complex survey data, and some forms of analysis may not be available!
In the DHS Recode manual, the sampling information for the data is found in variables v021 and v022, which are the primary sampling unit (PSU) and sample strata, respectively. The person weight is found in variable v005, and following DHS protocol, this has six implied decimal places, so we must divide it by 1000000, again, following the DHS manual.
library(survey)## Loading required package: grid## Loading required package: Matrix##
## Attaching package: 'survey'## The following object is masked from 'package:graphics':
##
## dotchartmodel.dat$wt<-model.dat$v005/1000000
#create the design: ids == PSU, strata==strata, weights==weights.
options(survey.lonely.psu = "adjust")
des<-svydesign(ids=~v021, strata = ~v022, weights=~wt, data=model.dat)
fit.s<-svykm(Surv(death.age, d.event)~rural, design=des, se=T)
#use svyby to find the %of infants that die before age 1, by rural/urban status
svyby(~d.event, ~rural, des, svymean)The plotting is a bit more of a challenge, as the survey version of the function isn’t as nice
plot(fit.s[[2]], ylim=c(.8,1), xlim=c(0,12),col=1, ci=F )
lines(fit.s[[1]], col=2)
title(main="Survival Function for Infant Mortality", sub="Rural vs Urban Residence")
legend("topright", legend = c("Urban","Rural" ), col=c(1,2), lty=1)
#test statistic
svylogrank(Surv(death.age, d.event)~rural, design=des)## Warning in regularize.values(x, y, ties, missing(ties), na.rm = na.rm):
## collapsing to unique 'x' values## [[1]]
## score
## [1,] 26.88143 15.11796 1.778112 0.07538546
##
## [[2]]
## chisq p
## 3.16168231 0.07538546
##
## attr(,"class")
## [1] "svylogrank"And we see the p-value is larger than assuming random sampling.